Cao + H2O ---->Ca(OH)2
Calculate the number of each reactant and the moles of the product
that is
moles = mass/molar mass
The moles of CaO= 56.08g/ 56.08g/mol(molar mass of Cao)= 1mole
the moles of water= 36.04 g/18 g/mol= 2.002moles
The moles of Ca (OH)2=74.10g/74.093g/mol= 1mole
The mass of differences of reactant and product can be therefore
explained as
1 mole of Cao reacted completely with 1 mole H2O to produce 1 mole of Ca(OH)2. The mass of water was in excess while that of CaO was limited
Answer:
63.05% of MgCO3.3H2O by mass
Explanation:
<em>of MgCO3.3H2O in the mixture?</em>
The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:
<em>Mass water:</em>
3.883g - 2.927g = 0.956g water
<em>Moles water -18.01g/mol-</em>
0.956g water * (1mol/18.01g) = 0.05308 moles H2O.
<em>Moles MgCO3.3H2O:</em>
0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =
0.01769 moles MgCO3.3H2O
<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>
0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O
<em>Mass percent:</em>
2.448g MgCO3.3H2O / 3.883g Mixture * 100 =
<h3>63.05% of MgCO3.3H2O by mass</h3>
Answer:
And we have to calculate the number of moles of sucrose present in a lb mass of sucrose: Moles of sucrose=454⋅g342.30⋅g⋅mol−1=1.33⋅mol .
Explanation:
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